//https://leetcode.cn/problems/number-of-islands/

class Solution
{
    int dx[4] = {1, -1, 0, 0};
    int dy[4] = {0, 0, 1, -1};
    bool vis[301][301];
    int n, m;

public:
    int numIslands(vector<vector<char>> &grid)
    {
        n = grid.size(), m = grid[0].size();
        int ret = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                // 进入深搜，说明区域加1，将当前区域排查
                if (grid[i][j] == '1' && vis[i][j] == false)
                {
                    ++ret;
                    bfs(grid, i, j);
                }
            }
        }
        return ret;
    }
    // 从i，j位置开始宽搜,将当前区域所有为1的位置，vis标记为true
    void bfs(vector<vector<char>> &grid, int i, int j)
    {
        queue<pair<int, int>> q;
        q.push({i, j});
        vis[i][j] = true;
        while (q.size())
        {
            auto [a, b] = q.front();
            q.pop();
            for (int k = 0; k < 4; k++)
            {
                int x = a + dx[k], y = b + dy[k];
                if (x >= 0 && x < n && y >= 0 && y < m && grid[x][y] == '1' && vis[x][y] == false)
                {
                    q.push({x, y});
                    vis[x][y] = true;
                }
            }
        }
    }
};